Problem: As you very well know, this year's funkiest numbers are so called triangular numbers (that is, integers that are representable as <image>, where k is some positive integer), and the coolest numbers are those that are representable as a sum of two triangular numbers.

A well-known hipster Andrew adores everything funky and cool but unfortunately, he isn't good at maths. Given number n, help him define whether this number can be represented by a sum of two triangular numbers (not necessarily different)!

Input

The first input line contains an integer n (1 ≤ n ≤ 109).

Output

Print "YES" (without the quotes), if n can be represented as a sum of two triangular numbers, otherwise print "NO" (without the quotes).

Examples

Input

256


Output

YES


Input

512


Output

NO

Note

In the first sample number <image>.

In the second sample number 512 can not be represented as a sum of two triangular numbers.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long a[52000];
void init() {
  int i = 0;
  int k;
  k = 1;
  while (1) {
    i++;
    if ((i + 1) * i >= 2100000000) break;
    k = i * (i + 1);
    a[i] = k / 2;
  }
}
int got(int x) {
  int l, r, m;
  l = 1;
  r = 45825;
  m = (l + r) >> 1;
  while (1) {
    if (x == 997146153) {
      l = l;
      r = r;
    }
    if (a[m] == x || a[r] == x || a[l] == x) return 1;
    if (x > a[m])
      l = m;
    else
      r = m;
    m = (l + r) >> 1;
    if (a[l] != x && a[r] != x && (r - l <= 1)) return 0;
  }
}
int main() {
  int t, l, r, m, i, n, flag = 0;
  init();
  scanf("%d", &n);
  l = 1;
  r = 45825;
  for (i = 1; i <= 45825; i++) {
    m = n - a[i];
    if (a[i] > n) break;
    if (got(m)) {
      flag = 1;
      break;
    }
  }
  if (flag)
    printf("YES\n");
  else
    printf("NO\n");
}